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Cannibals and Missionaries
Posted 09 August 2007 - 02:56 AM
how says you cant put a rope on the boat so you can send a missionarie and a cannibal to cross. then the Cannibals and Missionaries on the other side pull the boat back and then the next missionarie and cannibal go. then the last missionarie and cannibal but the rope in the boat and cross the river there you go.
Posted 14 August 2007 - 09:09 PM
1m and 1 c
m go back
1m and 1 c
c go back
1m and 1c
c go back
Posted 29 August 2007 - 06:17 PM
Once again, C=cannibal
M + M go over. M comes back, picks up another M brings it over, then comes back again to pick up C, brings him over. C comes back over and picks up another C, drops him off and comes back again with the last C.
This one fails right off the bat. As soon as the two missionaries go over, one missionary is left w/ three cannibals.
Posted 29 August 2007 - 06:32 PM
Agreed. The only way this works is assuming that the cannibals are only cannibals while on shore, not in the boat. If they maintained their cannibal tendencies in the boat, here is how the given answer would break down...
i think admin is wrong with taking 2 cannibals over at the same time and expecting on to return, if there is a m on the other side i think that both c's will stay and ,,,,well do what they do.
Cannibals and Missionaries - solution
1 cannibal and 1 missionary there, missionary back. 2 cannibals there, all three cannibals stay and eat lone missionary.
Posted 31 August 2007 - 01:42 AM
Posted 19 October 2007 - 05:39 PM
/x/ = River
THIS IS THE ANSWER. READ CAREFULLY AND YOU WILL UNDERSTAND
NOBODY ON ONE SIDE, NOBODY IN RIVER, ALL ON ONE SIDE
TWO CANNIBALS GO
ONE CANNIBAL DROPPED OFF, ONE GOES BACK
CANNIBAL ON ONE SIDE.CANNIBAL AND MISSIONARY IN RIVER, TWO MISSIONARY AND ONE CANNIBAL OTHER SIDE
C&M ONE SIDE, C & M IN R, C & M OTHER SIDE
C & TWO M ONE SIDE, C IN R, C & M OTHER SIDE
C & TWO M ONE SIDE, C & M IN R, C OTHER SIDE
C& THREE M ONE SIDE, C IN R, C OTHER SIDE
C & 3 M ONE SIDE, TWO C IN R, NOBODY OTHER SIDE
3 C & 3 M ONE SIDE, NOBODY IN R, NOBODY OTHER SIDE
WAMMY! THAT'S IT!
Posted 14 November 2007 - 03:08 AM
Missionary = YOU
You and i cross the river leaving two of me and two of you. I drop you off on the other side of the river. I go back and pick up a another me, leaving two of you and one of me. I drop myself off. Leaving only one of each of us. I go back and pick up you. We start crossing, leaving one me and two you. I drop you off and go back. i pick myself up and cross the river dropping myself off. there are two me and two you. I go back and pick you up the last time. no one is left. We get across and get off. leaving the boat... everyone there...!
Posted 30 November 2007 - 10:17 PM
C/CM/MMC < once the CM arives <- you have 2 CC to 1 M = dead/eaten M
Posted 30 November 2007 - 10:28 PM
MMMCCC | | MMMC | |CC - CC cross the river MMMCC | |C - C goes back MMM | |CCC - CC cross the river MMMC | |CC - C goes back M C | |MMCC - MM cross the river MM CC | |M C - MC goes back CC | |MMMC - MM cross the river CCC | |MMM - C goes back C | |MMMCC - CC cross the river CC | |MMMC - C goes back | |MMMCCC - CC cross the river
Posted 07 December 2007 - 07:27 AM
Just kidding! I like the first solution. As for the question of wording, the riddle states that the point of not having the cannibals outnumber the missionaries is so the cannibals don't eat the missionaries. Therefore it's obvious that having 1 cannibal and 0 missionaries is safe!
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